the position of a body moving along x-axis at time t is given by x = (t2-4t+6) m . The distance travelled by body in time interval t = 0 to t = 3 s is -
(1) 5 m
(2) 7m
(3) 4 m
(4) 3 m
Here x=t^2-4t+6 equation represent the position of that particle along x axis at given time t. Now at t=0 , particle present at x=6 point means x(6,0) then t=1 ,x(3,1); then t=2,x(2,2); and last t=3,x(3,3). Now distance is sclar quantity so along x axis particle move the distance x(6,0) to x(3,1) = 3m, x(3,1) to x(2,2) =1m and last the particle move x(2,2) to x(3,1)=1m . So the distance travelled by body in time interval t=0 to t=3s is (3+1+1)=5m
- 74
As already told by Abu syed.
x at 0 sec = 0^2-4*0+6=6m
x at 1 sec = 3m
x at 2 sec = 2m
x at 3 sec = 3m
Now let's take out distance at particular second.
x at 1st sec = 3m
x at 2nd sec (x at 2 sec - x at 1sec) = 2-3 = 1m (as distance is scalar quantity and hence has no direction therefore no-ve sign)
x at 3rd sec (x at 3 sec - x at 2sec) = 3-2 = 1m.
Now add all the individual distances I.e. 3+1+1 = 5m.
- 110
Abu syed. x at 0 sec = 0^2-4*0+6=6m x at 1 sec = 3m x at 2 sec = 2m x at 3 sec = 3m Now let's take out distance at particular second. x at 1st sec = 3m x at 2nd sec (x at 2 sec - x at 1sec) = 2-3 = 1m (as distance is scalar quantity and hence has no direction therefore no-ve sign) x at 3rd sec (x at 3 sec - x at 2sec) = 3-2 = 1m. Now add all the individual distances I.e. 3+1+1 = 5m.
- 0
X =t - 4 + t b First we have to check where velocity of particle is change v= DX DT = 2+-4=2 (t-2) velocity is changing at t 2 sec motion of particle is like at t=0= X= 6so , magnitude of distance travelled (6-2)=4m at +=3=X=3 so distance travelled during t = 2sec of =X3 -x 2=3-2 =1m so Total distance travelled =(4+1)m = 5m
- 1
The correct option is A.
Given,
x=t2?4t+6
So, the change in velocity is:
dtdx?=2t?4
Since velocity is changing,
At?t=0,x1?=6
At?t=2,x2?=2
So the magnitude of the distance traveled is:
6?2=4m
At?t=3,x3?=3
So distance traveled from?t=2s?to?t=3s
x3??x2?=3?2=1m
Thus the total distance is:
4+1=5m
Given,
x=t2?4t+6
So, the change in velocity is:
dtdx?=2t?4
Since velocity is changing,
At?t=0,x1?=6
At?t=2,x2?=2
So the magnitude of the distance traveled is:
6?2=4m
At?t=3,x3?=3
So distance traveled from?t=2s?to?t=3s
x3??x2?=3?2=1m
Thus the total distance is:
4+1=5m
- 0